Answer:
ΔH = 
Explanation:
In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.
The heat transfer is represented by
=
where
= the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.
= the heat of combustion
Also, we know that the total heat change of the any system is
ΔH = ΔQ + ΔW
where
ΔH = the total heat absorbed by the system
ΔQ = the internal heat absorbed by the system which in this case is
ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0
substituting into the heat change equation
ΔH = + 0
==> ΔH =
Answer:
Volume in L = 0.50 L
Explanation:
Given data:
Molarity of solution = 0.1850 M
Mass of BaCl₂ = 19.30 g
Volume of solution = ?
Solution:
First of all we will calculate the number of moles of solute:
Number of moles = mass/molar mass
Molar mass of BaCl₂ = 208.23 g/mol
Number of moles = 19.30 g / 208.23 g/mol
Number of moles = 0.093 mol
Volume of solution:
Molarity = number of moles of solute / volume of solution in L
0.1850 M = 0.093 mol / Volume in L
Volume in L = 0.093 mol / 0.1850 M (M= mol/L)
Volume in L = 0.50 L
Answer:
it holds reactant molecules in a good position for then to react
Explanation:
The pressure exerted by vapors or gas on the surface of a liquid is known as vapor pressure.
This means that weaker is the intermolecular forces present in a substance more easily it can form vapors. As a result, it will have high vapor pressure.
As substance B has high vapor pressure which means that it has weak intermolecular forces.
Also, stronger is the intermolecular forces present in a substance more will be its boiling point. Hence, more energy or temperature is required to break the bonds. Hence, substance A has higher boiling point and high heat of vaporization.
When surrounding pressure is less than or equal to its vapor pressure then substance B boils into the gas phase. Hence, substance B will be a gas at 300 mm Hg.
Therefore, we can conclude that characteristics of the two substances will be as follows.
(a) Substance B - has weaker intermoclcular
(b) Substance A - has a higher boiling point
(c) Substance B - is a gas at 300 mm Hg
(d) Substance A - has a higher heat of vaporization
